\(I=\int_{1}^{e}\frac{1}{x}(x^2+3\sqrt{1+3lnx})dx=\int_{1}^{e}xdx+\int_{1}^{e}\frac{3}{x}\sqrt{1+3lnx}dx=A+B\) \(A=\int_{1}^{e}xdx=\frac{x^2}{2}\bigg |^e_1=\frac{e^2-1}{2}\) Tính \(B=\int_{1}^{e}\frac{3}{x}\sqrt{1+3lnxdx}.\) Đặt \(t=\sqrt{1+3lnx}\Rightarrow t^2=1+3lnx\Rightarrow 2tdt=\frac{3}{x}dx\) Đổi cận Suy ra \(B=\int_{1}^{2}t.2tdt=2\int_{1}^{2}t^2dt=\frac{2t^3}{3}\bigg |^2_1=\frac{14}{1}\) Vậy \(I=A+B=\frac{e^2}{2}+\frac{25}{6}\)